SOLUTION: what is the center and radius of the equation below? x^2+y^2-8x-2y+16=0

Question 554129: what is the center and radius of the equation below?
x^2+y^2-8x-2y+16=0
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
Complete the square on this one. First, get the x terms together and the y terms together, leaving a blank to add to the x terms and y terms in order to complete the square. Place similar blanks on the right side so you can add numbers to BOTH sides of the equation, as follows:

x^2 -8x + ____ + y^2 -2y + _____ = -16 + ____ + ____

Take half of the -8, which is -4, and square, which would be 16.
Take half of the -2, which is -1, and square, which would be 1. Fill in the blanks above with these numbers, adding the numbers to each side of the equation.

x^2 - 8x + 16 + y^2 - 2y + 1 = -16 +16 +1
(x-4)^2 + (y-1)^2 = 1

Therefore the center is at (4,1), and the r^2=1, so the radius is 1.

You may want to see my FREE website for a non-traditional explanation of this topic. The easiest way to find the website is to use the easy-to-remember and easy-to-spell link www.mathinlivingcolor.com. Near the bottom of this page is a link that takes you to my Homepage.

On my Homepage, look for the link "Basic, Intermediate, and College Algebra: One Step at a Time." Choose "College Algebra" and look in "Chapter 2" for "Section 2.04 Circles." I think you will really like the "Math in Living Color" pages that go with this section.

If you need to contact me, send me an Email at rapaljer@seminolestate.edu.

Dr. Robert J. Rapalje, Retired
Seminole State College of Florida
Altamonte Springs Campus

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